A100320 -id:A100320 - cp's OEIS frontend

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
  rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
  For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

T. D. Noe, Rows n=0..50 of triangle, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Quang T. Bach and Jeffrey B. Remmel, Generating functions for descents over permutations which avoid sets of consecutive patterns, arXiv:1510.04319 [math.CO], 2015 (see p.25).
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Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
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Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
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Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 15, 29.
Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 12.
Jonathan E. Beagley and Paul Drube, Combinatorics of Tableau Inversions, Electron. J. Combin., 22 (2015), #P2.44.
S. Chakravarty and Y. Kodama, A generating function for the N-soliton solutions of the Kadomtsev-Petviashvili II equation, arXiv preprint arXiv:0802.0524v2 [nlin.SI], 2008.
Wun-Seng Chou, Tian-Xiao He, and Peter J.-S. Shiue, On the Primality of the Generalized Fuss-Catalan Numbers, J. Int. Seqs., Vol. 21 (2018), #18.2.1.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 11.
Paul Drube, Generating Functions for Inverted Semistandard Young Tableaux and Generalized Ballot Numbers, arXiv:1606.04869 [math.CO], 2016.
Paul Drube, Generalized Path Pairs and Fuss-Catalan Triangles, arXiv:2007.01892 [math.CO], 2020. See Figure 4 p. 8.
T.-X. He and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, example p 32.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Thomas Koshy, Lobb's generalization of Catalan's parenthesization problem, The College Mathematics Journal 40 (2), March 2009, 99-107, DOI:10.1080/07468342.2009.11922344.
Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 11.
Andrew Lobb, Deriving the n-th Catalan number, Mathematical Gazette, Vol. 83, No. 496 (March 1999), 109-110.
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Wikipedia, Lobb number
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Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

/* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015

T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023

Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)

a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
trianglerows(10) \\ Felix Fröhlich, Jun 24 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle
def A039599_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True ; h = 1
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        if b : print([D[z] for z in (1..h-1)])
        b = not b
A039599_triangle(10)  # Peter Luschny, May 01 2012

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 5, 3, 1, 0, 14, 14, 9, 4, 1, 0, 42, 42, 28, 14, 5, 1, 0, 132, 132, 90, 48, 20, 6, 1, 0, 429, 429, 297, 165, 75, 27, 7, 1, 0, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 0, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1

Offset: 0

Philippe Deléham, May 30 2005

Catalan convolution triangle; g.f. for column k: (x*c(x))^k with c(x) g.f. for A000108 (Catalan numbers).
Riordan array (1, xc(x)), where c(x) the g.f. of A000108; inverse of Riordan array (1, x*(1-x)) (see A109466).
Diagonal sums give A132364. - Philippe Deléham, Nov 11 2007

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,  1;
  0,   5,   5,  3,  1;
  0,  14,  14,  9,  4,  1;
  0,  42,  42, 28, 14,  5, 1;
  0, 132, 132, 90, 48, 20, 6, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)

Alois P. Heinz, Rows n = 0..140, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
F. R. Bernhart, Catalan, Motzkin and Riordan numbers, Discr. Math., 204 (1999), 73-112.
D. Callan, A recursive bijective approach to counting permutations containing 3-letter patterns, arXiv:math/0211380 [math.CO], 2002.
E. Deutsch, Dyck path enumeration, Discrete Math., 204, 1999, 167-202.
FindStat - Combinatorial Statistic Finder, The number of touch points of a Dyck path, The number of initial rises of a Dyck paths, The number of nodes on the left branch of the tree, The number of subtrees.
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
A. Robertson, D. Saracino and D. Zeilberger, Refined restricted permutations, arXiv:math/0203033 [math.CO], 2002.
L. W. Shapiro, S. Getu, W.-J. Woan and L. C. Woodson, The Riordan group, Discrete Applied Math., 34 (1991), 229-239.

Column k for k = 0, 1, 2, ..., 13: A000007, A000108, A000108, A000245, A002057, A000344, A003517, A000588, A003517, A001392, A003518, A000589, A003519, A000590
The three triangles A059365, A106566 and A099039 are the same except for signs and the leading term.
Diagonals: A000012, A001477, A000096, A005586, A005587, A005557, A064059, A064061
See also A009766, A033184, A059365 for other versions.
Generalized Catalan numbers C(x, n) for -11 <= x <= 10: A064333, A064332, A064331, A064330, A064329, A064328, A064327, A064326, A064325, A064311, A064310, A000012, A000108, A064062, A064063, A064087, A064088, A064089, A064090, A064091, A064092, A064093.
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Diagonals give A000108 A000245 A002057 A000344 A003517 A000588 A003518 A003519 A001392, ...

A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >;
[A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021

A106566 := proc(n,k)
    if n = 0 then
        1;
    elif k < 0 or k > n then
        0;
    else
        binomial(2*n-k-1,n-k)*k/n ;
    end if;
end proc: # R. J. Mathar, Mar 01 2015

T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *)
(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

{T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */

def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k)
flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021

T(n, k) = binomial(2n-k-1, n-k)*k/n for 0 <= k <= n with n > 0; T(0, 0) = 1; T(0, k) = 0 if k > 0.
T(0, 0) = 1; T(n, 0) = 0 if n > 0; T(0, k) = 0 if k > 0; for k > 0 and n > 0: T(n, k) = Sum_{j>=0} T(n-1, k-1+j).
Sum_{j>=0} T(n+j, 2j) = binomial(2n-1, n), n > 0.
Sum_{j>=0} T(n+j, 2j+1) = binomial(2n-2, n-1), n > 0.
Sum_{k>=0} (-1)^(n+k)*T(n, k) = A064310(n). T(n, k) = (-1)^(n+k)*A099039(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = 0,1,2,3,4,5,6,7,8 respectively.
Sum_{k>=0} T(n, k)*x^(n-k) = C(x, n); C(x, n) are the generalized Catalan numbers.
Sum_{j=0..n-k} T(n+k,2*k+j) = A039599(n,k).
Sum_{j>=0} T(n,j)*binomial(j,k) = A039599(n,k).
Sum_{k=0..n} T(n,k)*A000108(k) = A127632(n).
Sum_{k=0..n} T(n,k)*(x+1)^k*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x= 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Aug 25 2007
Sum_{k=0..n} T(n,k)*A000108(k-1) = A121988(n), with A000108(-1)=0. - Philippe Deléham, Aug 27 2007
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 27 2007
T(n,k)*2^(n-k) = A110510(n,k); T(n,k)*3^(n-k) = A110518(n,k). - Philippe Deléham, Nov 11 2007
Sum_{k=0..n} T(n,k)*A000045(k) = A109262(n), A000045: Fibonacci numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A000129(k) = A143464(n), A000129: Pell numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A100335(k) = A002450(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A100334(k) = A001906(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A099322(k) = A015565(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A106233(k) = A003462(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A151821(k+1) = A100320(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A082505(k+1) = A144706(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A000045(2k+2) = A026671(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A122367(k) = A026726(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A008619(k) = A000958(n+1). - Philippe Deléham, Nov 15 2009
Sum_{k=0..n} T(n,k)*A027941(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
G.f.: Sum_{n>=0, k>=0} T(n, k)*x^k*z^n = 1/(1 - x*z*c(z)) where c(z) the g.f. of A000108. - Michael Somos, Oct 01 2022

Formula corrected by Philippe Deléham, Oct 31 2008
Corrected by Philippe Deléham, Sep 17 2009
Corrected by Alois P. Heinz, Aug 02 2012

A028329 Twice central binomial coefficients.

Original entry on oeis.org

2, 4, 12, 40, 140, 504, 1848, 6864, 25740, 97240, 369512, 1410864, 5408312, 20801200, 80233200, 310235040, 1202160780, 4667212440, 18150270600, 70690527600, 275693057640, 1076515748880, 4208197927440, 16466861455200, 64495207366200, 252821212875504, 991837065896208

Offset: 0

Mohammad K. Azarian

Central elements in the even-Pascal triangle A028326.
If Y is a 3-subset of an 2n-set X then, for n>=3, a(n-1) is the number of (n+1)-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
a(n) denotes the number of ways one can reach the (n,n) point in an n X n grid via the point (n-1, n-1) starting from (0,0) when moving right and up is allowed [From Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 29 2009]
It appears that a(n-1) is also the number of quivers in the mutation class of twisted types BD_n and CD_n for n >= 3. - Christian Stump, Nov 03 2010
This is the case m = n+1 in the Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
From Ran Pan, Feb 01 2016: (Start)
a(n) is the number of North-East paths from (0,0) to (n+1,n+1) that bounce off the diagonal y = x an even number of times. Details can be found in Section 4.2 in Pan and Remmel's link.
a(n) is the number of North-East paths from (0,0) to (n+1,n+1) that cross the diagonal y = x an even number of times. Details can be found in Section 4.3 in Pan and Remmel's link. (End)

Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Anwar Al Ghabra, K. Gopala Krishna, Patrick Labelle, and Vasilisa Shramchenko, Enumeration of multi-rooted plane trees, arXiv:2301.09765 [math.CO], 2023.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Ran Pan and Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.

Bisection of A047073, A063886.
First differences of A054113.
Cf. A000984, A095660, A100320, A162551.
Cf. A028326, A028327, A028328, A028330, A028331, A028332.

[2*(n+1)*Catalan(n): n in [0..30]]; // G. C. Greubel, Jul 13 2024

seq(add(binomial(2*n,n),k=1..2),n=0..23); # Zerinvary Lajos, Dec 14 2007

Table[2Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Aug 08 2011 *)

PARI
```
a(n)=2*binomial(2*n,n)
```

[2*binomial(2*n,n) for n in range(31)] # G. C. Greubel, Jul 13 2024

G.f.: 2/sqrt(1 - 4*x).
a(n) = 2*A000984(n).
a(n) = 2 * binomial(2*n, n).
a(n) = A100320(n) = A095660(2*n,n) for n > 0. - Reinhard Zumkeller, Apr 08 2012
G.f.: G(0), where G(k)= 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) + (k + 1)/ G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 07 2013
a(n) = binomial(2*n+2, n+1) - A162551(n). - Ran Pan, Feb 01 2016
D-finite with recurrence: n*a(n) + 2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x). - Stefano Spezia, May 11 2024

Edited by Michael Somos, Sep 13 2003

A095660 Pascal (1,3) triangle.

Original entry on oeis.org

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3

Offset: 0

Wolfdieter Lang, May 21 2004

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012
Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

			Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
Cf. A097805.

a095660 n k = a095660_tabl !! n !! k
a095660_row n = a095660_tabl !! n
a095660_tabl = [3] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
-- Reinhard Zumkeller, Apr 08 2012

A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
[A095660(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021

T(n,k):=piecewise(n=0,3,0Mircea Merca, Apr 08 2012

{3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)

def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

A132046 Triangle read by rows: T(n,0) = T(n,n) = 1, and T(n,k) = 2*binomial(n,k) for 1 <= k <= n - 1.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 6, 6, 1, 1, 8, 12, 8, 1, 1, 10, 20, 20, 10, 1, 1, 12, 30, 40, 30, 12, 1, 1, 14, 42, 70, 70, 42, 14, 1, 1, 16, 56, 112, 140, 112, 56, 16, 1, 1, 18, 72, 168, 252, 252, 168, 72, 18, 1, 1, 20, 90, 240, 420, 504, 420, 240, 90, 20, 1

Offset: 0

Gary W. Adamson, Aug 08 2007

T(2*n,n) is A100320 (with Hankel transform A144704). - Paul Barry, Sep 19 2008
Double the internal elements of Pascal's triangle. - Paul Barry, Jan 07 2009
Coefficients of 2*(x + 1)^n - (x^n + 1) as a triangle (except for the very first term). - Thomas Baruchel, Jun 02 2018

			First few rows of the triangle are:
  1;
  1,  1;
  1,  4,  1;
  1,  6,  6,  1;
  1,  8, 12,  8,  1;
  1, 10, 20, 20, 10,  1;
  1, 12, 30, 40, 30, 12,  1;
  1, 14, 42, 70, 70, 42, 14, 1;
  ...

Row sums: A095121.
Cf. A007318, A103451, A132046.
Cf. A154327 (diagonal sums). [Paul Barry, Jan 07 2009]
Cf. A141540.

T[n_, k_] := If[n == k || k == 0, 1, If[k <= n, 2 Binomial[n, k], 0]]
Flatten[Table[T[n, k], {n, 0, 20}, {k, 0, n}]] (* Emanuele Munarini, May 15 2018 *)

T(n, k) := if k = 0 or k = n then 1 else 2*binomial(n, k)$
create_list(T(n, k), n, 0, 20, k, 0, n); /* Franck Maminirina Ramaharo, Jan 03 2019 */

T(n,k) = 2*A007318(n,k) - A103451(n,k).
T(n,k) = [k<=n] (0^(n + k) + C(n,k)*(2 - 0^(n - k) - 0^k)). - Paul Barry, Sep 19 2008
T(n,k) = A007318(n,k)*A154325(n,k). - Paul Barry, Jan 07 2009
From Emanuele Munarini, May 15 2018: (Start)
G.f.: (1 - t - x*t + 3*x*t^2 - x*t^3 - x^2*t^3)/((1 - t)*(1 - x*t)*(1 - t - x*t)).
T(n+3,k+2) = 2*T(n+2,k+2) - T(n+1,k+2) + 2*T(n+2,k+1) - 3*T(n+1,k+1) - T(n+1,k) + T(n,k+1) + T(n,k), except for n = 0 and k = 0. (End)
E.g.f.: 1 - exp(t) - exp(t*x) + 2*exp(t*(1 + x)). - Franck Maminirina Ramaharo, Jan 02 2019

A124927 Triangle read by rows: T(n,0)=1, T(n,k)=2*binomial(n,k) if k>0 (0<=k<=n).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 6, 2, 1, 8, 12, 8, 2, 1, 10, 20, 20, 10, 2, 1, 12, 30, 40, 30, 12, 2, 1, 14, 42, 70, 70, 42, 14, 2, 1, 16, 56, 112, 140, 112, 56, 16, 2, 1, 18, 72, 168, 252, 252, 168, 72, 18, 2, 1, 20, 90, 240, 420, 504, 420, 240, 90, 20, 2, 1, 22, 110, 330, 660, 924, 924, 660, 330, 110, 22, 2

Offset: 0

Gary W. Adamson, Nov 12 2006

Pascal triangle with all entries doubled except for the first entry in each row. A028326 with first column replaced by 1's. Row sums are 2^(n+1)-1.
From Paul Barry, Sep 19 2008: (Start)
Reversal of A129994. Diagonal sums are A001595. T(2n,n) is A100320.
Binomial transform of matrix with 1,2,2,2,... on main diagonal, zero elsewhere. (End)
This sequence is jointly generated with A210042 as an array of coefficients of polynomials v(n,x):  initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+v(n-1,x) +1 and v(n,x)=x*u(n-1,x)+x*v(n-1,x).  See the Mathematica section. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012

			Triangle starts:
  1;
  1,  2;
  1,  4,  2;
  1,  6,  6,  2;
  1,  8, 12,  8,  2;
  1, 10, 20, 20, 10, 2;
(1, 0, 0, 1, 0, 0, ...) DELTA (0, 2, -1, 0, 0, ...) begins:
  1;
  1,  0;
  1,  2,  0;
  1,  4,  2,  0;
  1,  6,  6,  2,  0;
  1,  8, 12,  8,  2, 0;
  1, 10, 20, 20, 10, 2, 0. - _Philippe Deléham_, Mar 25 2012

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000225.

Cf. A074909.

a124927 n k = a124927_tabl !! n !! k
a124927_row n = a124927_tabl !! n
a124927_tabl = iterate
   (\row -> zipWith (+) ([0] ++ reverse row) (row ++ [1])) [1]
-- Reinhard Zumkeller, Mar 04 2012

[k eq 0 select 1 else 2*Binomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 10 2019

T:=proc(n,k) if k=0 then 1 else 2*binomial(n,k) fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210042 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A124927 *) (* Clark Kimberling, Mar 17 2012 *)
(* Second program *)
Table[If[k==0, 1, 2*Binomial[n, k]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 10 2019 *)

T(n,k) = if(k==0,1, 2*binomial(n,k)); \\ G. C. Greubel, Jul 10 2019

def T(n, k):
    if (k==0): return 1
    else: return 2*binomial(n,k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 10 2019

T(n,0) = 1; for n>0: T(n,n) = 2, T(n,k) = T(n-1,k) + T(n-1,n-k), 1Reinhard Zumkeller, Mar 04 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2012
G.f.: (1-x+x*y)/((-1+x)*(x*y+x-1)). - R. J. Mathar, Aug 11 2015

Edited by N. J. A. Sloane, Nov 24 2006

cp's OEIS Frontend

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

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A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

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A028329 Twice central binomial coefficients.

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A095660 Pascal (1,3) triangle.

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A132046 Triangle read by rows: T(n,0) = T(n,n) = 1, and T(n,k) = 2*binomial(n,k) for 1 <= k <= n - 1.

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A124927 Triangle read by rows: T(n,0)=1, T(n,k)=2*binomial(n,k) if k>0 (0<=k<=n).

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A144704 a(n) = 4^n(1-2n).

A146534 a(n) = 4C(2n,n) - 30^n.